A capacitor is constructed of two aluminum plate separated by a dielectric. the capacitance is measured to be 268.00 uf. without changing the dimensions of the plates, they are brought closer together so that the distance between the two plates is 1/2 of the original distance.
What is the new capacitance (in μF)?
Twice
see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
same E, less distance
typo - excuse me, same V, less distance, so greater E
@Damon, can you explain me what you wrote
I don't understand
To find the new capacitance, we need to use the formula for the capacitance of a parallel plate capacitor:
C = (ε0 * εr * A) / d,
where:
C is the capacitance,
ε0 is the permittivity of free space (8.85 x 10^-12 F/m),
εr is the relative permittivity of the dielectric material (dimensionless),
A is the area of the plate,
d is the separation distance between the plates.
Given that the original capacitance (C1) is 268.00 μF and the new separation distance (d2) is half the original distance (d1), we can use this information to find the new capacitance (C2).
C1 = (ε0 * εr * A) / d1,
C2 = (ε0 * εr * A) / d2.
Dividing the first equation by the second equation, we get:
C1 / C2 = (d2 / d1).
Simplifying further:
C2 = C1 * (d1 / d2).
Substituting the given values:
C2 = 268.00 μF * (d1 / (d1 / 2)).
Since d1 cancels out:
C2 = 268.00 μF * 2.
C2 = 536.00 μF.
Therefore, the new capacitance (C2) is 536.00 μF.